Bashyboy said:Homework Statement
Find the sum of the series and its radius of convergence:
[itex]\sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x-1)^n}{n}[/itex]
Homework Equations
The Attempt at a Solution
I found the radius of convergence, but I wasn't sure how to find the sum of the power series.
Bashyboy said:Would it be [itex]f'(x) = \sum_{n=0}^{\infty} (-1)^{n+1}(x-1)^{n-1}[/itex]?
Bashyboy said:Would it be [itex]f'(x) = \sum_{n=0}^{\infty} (-1)^{n+1}(x-1)^{n-1}[/itex]?
Bashyboy said:If you distribute the (-1)^(n+1) to the (x-1)^(n-1) would it be a geometric series?
A power series is an infinite sum of terms, where each term is a polynomial function of the variable x. The general form of a power series is ∑(an)(xn), where an is a constant coefficient and n is a non-negative integer.
To find a function from a power series, you can use the process of power series expansion. This involves substituting the values of x into the power series and simplifying the resulting expression. The resulting function will be the sum of all the simplified terms.
A Taylor series is a special type of power series, where the terms are specifically chosen to approximate a given function at a specific point. Power series, on the other hand, have more general terms and do not necessarily have to be centered at a specific point.
No, not all functions can be represented by a power series. A function must be analytic, meaning it must have a continuous derivative of all orders, for it to be represented by a power series.
You can use a power series to approximate a function by truncating the series to a certain number of terms. The more terms you include, the more accurate the approximation will be. This is useful for evaluating complex functions or finding values that are difficult to calculate directly.